Answer:
C. Angle of Attack.
Explanation:
The pilot must adjust the angle of attack parameter. The angle of attack of this plane to get to the desired lift coefficient.
And thus, we have
Lift = Weight
Answer:
Velocity = 3.25[m/s]
Explanation:
This problem can be solved if we use the Bernoulli equation: In the attached image we can see the conditions of the water inside the container.
In point 1, (surface of the water) we have the atmospheric pressure and at point 2 the water is coming out also at atmospheric pressure, therefore this members in the Bernoulli equation could be cancelled.
The velocity in the point 1 is zero because we have this conditional statement "The water surface drops very slowly and its speed is approximately zero"
h2 is located at point 2 and it will be zero.
![(P_{1} +\frac{v_{1}^{2} }{2g} +h_{1} )=(P_{2} +\frac{v_{2}^{2} }{2g} +h_{2} )\\P_{1} =P_{2} \\v_{1}=0\\h_{2} =0\\v_{2}=\sqrt{0.54*9.81*2}\\v_{2}=3.25[m/s]](https://tex.z-dn.net/?f=%28P_%7B1%7D%20%2B%5Cfrac%7Bv_%7B1%7D%5E%7B2%7D%20%7D%7B2g%7D%20%2Bh_%7B1%7D%20%29%3D%28P_%7B2%7D%20%2B%5Cfrac%7Bv_%7B2%7D%5E%7B2%7D%20%7D%7B2g%7D%20%2Bh_%7B2%7D%20%29%5C%5CP_%7B1%7D%20%3DP_%7B2%7D%20%5C%5Cv_%7B1%7D%3D0%5C%5Ch_%7B2%7D%20%3D0%5C%5Cv_%7B2%7D%3D%5Csqrt%7B0.54%2A9.81%2A2%7D%5C%5Cv_%7B2%7D%3D3.25%5Bm%2Fs%5D)
Effort force and Resistance force
Answer:
Beats are the difference in frequency.
(a) is correct option.
Explanation:
Beat :
Beat is the difference of the frequency of two waves.
The difference in frequency is equal to the number of beat per second.
Amplitude :
Amplitude of the wave is the maximum displacement.
Frequency :
Frequency is the number oscillations of wave in per second.
Intensity :
Intensity is the power per unit area.
Hence, Beats are the difference in frequency.
Answer:
A) M
Explanation:
The three blocks are set in series on a horizontal frictionless surface, whose mutual contact accelerates all system to the same value due to internal forces as response to external force exerted on the box of mass M (Newton's Third Law). Let be F the external force, and F' and F'' the internal forces between boxes of masses M and 2M, as well as between boxes of masses 2M and 3M. The equations of equilibrium of each box are described below:
Box with mass M

Box with mass 2M

Box with mass 3M

On the third equation, acceleration can be modelled in terms of F'':

An expression for F' can be deducted from the second equation by replacing F'' and clearing the respective variable.



Finally, F'' can be calculated in terms of the external force by replacing F' on the first equation:




Afterwards, F' as function of the external force can be obtained by direct substitution:

The net forces of each block are now calculated:
Box with mass M


Box with mass 2M


Box with mass 3M

As a conclusion, the box with mass M experiments the smallest net force acting on it, which corresponds with answer A.